$\int (-3 x^4 -6 x^2 +8)\,dx=$ $+C$
Solution: We can use the sum rule and the constant multiple rule for indefinite integrals: $\begin{aligned} &\int [f(x)+g(x)]dx=\int f(x)\,dx+\int g(x)\,dx \\\\\\ &\int k\cdot f(x)= k\cdot\int f(x)\,dx \end{aligned}$ Using the sum and the constant multiple rules, we can rewrite our integral as follows: $\int (-3 x^4 -6 x^2 +8)\,dx= -3\int x^4\,dx -6\int x^2\,dx +8\int 1\,dx$ Now we can find each indefinite integral using the reverse power rule: $\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C$ Note: we can only use the reverse power rule because $n \neq -1$. $\begin{aligned} &\phantom{=}\int (-3 x^4 -6 x^2 +8)\,dx \\\\ &= -3\int x^4\,dx -6\int x^2\,dx +8\int 1\,dx \\\\ &=-3 \dfrac{x^5}{5} -6\dfrac{x^3}{3} +8\dfrac{x^1}{1}+C \\\\ &=-\dfrac{3}{5} x^5 -2 x^3 +8 x+C \end{aligned}$ In conclusion, $\int (-3 x^4 -6 x^2 +8)\,dx=-\dfrac{3}{5} x^5 -2 x^3 +8 x+C$